ChemE 178:
Homework #6 Solutions
1.) Polymer blends are to be made from a poly(ethylenepropylene) (hPEP) of weight
average molecular weight Mw = 175,000, Mw/Mn=1.01 and a perdeuterated
poly(ethylenepropylene)
(dPEP), Mw=200,000, Mw/Mn=1.01. The dPEP is the
same in all respect as the hPEP except that the dPEP has all hydrogens
exchanged for deuterium. The Flory parameter
χ
for the dPEP/hPEP pair is
given by
χ
= (0.571/T)6.6x10
4
,
where T is the absolute temperature (K). Compute and plot the following:
_______________________________________________________________________
_
A.)
The normalized free energy of mixing
∆
G
mix
/Mk
B
T versus
ϕ
dPB
at the following
temperatures: 300K, 350K, 385K and 450K. Use a spreadsheet program like
Excel or a plotting program with computational capability to produce these plots.
Use scales so that important details are visible.
Begin with the FloryHuggins Equation:
(
29
(
29
ln
ln
A
B
A
B
A
B
B
A
B
G
Mk T
N
N
φ
φ
φ
φ
χφ φ
∆
=
+
+
To get N
A
and N
B
, you will need to use the number average molecular weight.
dPEP:
200,000
198,020
1.01
w
n
M
M
PDI
=
=
=
hPEP:
175,000
173,267
1.01
w
n
M
M
PDI
=
=
=
Assuming that the hPEP and dPEP have similar monomeric volumes, the
segment lengths are simply the degree of polymerization.
Note: This is not
always the case.
173,267
2,475
70
198,020
2,475
80
h
PEP
A
d
PEP
B
N
N
N
N


=
=
=
=
=
=
The polymer blend is incompressible, so
1
A
B
φ
φ
=
+
Given the expression for χ, you can use the FloryHuggins equation to plot the
normalized free energy as a function of volume fraction.
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Gibbs Energy of Mixing
1.40E04
1.20E04
1.00E04
8.00E05
6.00E05
4.00E05
2.00E05
0.00E+00
2.00E05
4.00E05
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
φ
(dPB)
∆
G/MkT
300 K
350 K
385 K
450 K
B.)
Compute the binodal and spinodal
ϕ
's as a function of temperature and use these
to compute and plot the phase diagram for this system.
The binodal is define as
(
29
(
29
(
29
1
2
1
1
ln
1
ln
1
1
2
A
A
mix
mix
A
B
A
A
A
A
B
G
G
N
N
φ
φ
φ
φ
χ
φ
φ
φ
∂∆
∂∆
=
=
+

+
+

÷
÷
∂
∂
The binodal is calculated using the condition that binodal point lies on the same
tangent line.
For this problem, the free energy curves are symmetric (N
A
= N
B
),
so the common tangent line is horizontal (a slope of zero).
As a result, the
binodal point is the local minimum of the free energy curve.
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 Fall '08
 SEGALMAN
 Thermodynamics, Entropy

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